Question: Rewrite the equation by completing the square. $2 x^{2} -11 x +14 = 0$ $(x + $
Solution: $\begin{aligned} 2 x^2 -11 x +14&=0 \\\\ 2 x^2 -11 x &=-14 \\\\ x^2 -\dfrac{11}{2} x&=-7 \end{aligned}$ Now we want to complete $x^2 -\dfrac{11}{2} x$ into a perfect square. To do that, we should add $\left(\dfrac{{-\frac{11}{2}}}{2}\right)^2={\dfrac{121}{16}}$ to it: $x^2{-\dfrac{11}{2}}x + {\dfrac{121}{16}}=\left(x -\dfrac{11}{4} \right)^2$ $\begin{aligned} x^2 -\dfrac{11}{2} x&=-7 \\\\ x^2 -\dfrac{11}{2} x + {\dfrac{121}{16}}&=-7 + {\dfrac{121}{16}} \\\\ \left(x -\dfrac{11}{4} \right)^2&=\dfrac{9}{16} \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x -\dfrac{11}{4} \right)^2=\dfrac{9}{16}$